Simplify and expand the following expression: $ \dfrac{n - 3}{n - 7}-\dfrac{2n + 6}{n - 2} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(n - 7)(n - 2)$ Multiply the first term by $\dfrac{n - 2}{n - 2}$ $ \begin{align*} \dfrac{n - 3}{n - 7} \times \dfrac{n - 2}{n - 2} & = \dfrac{(n - 3)(n - 2)}{(n - 7)(n - 2)} \\ & = \dfrac{n^2 - 5n + 6}{(n - 7)(n - 2)}\end{align*} $ Multiply the second term by $\dfrac{n - 7}{n - 7}$ $ \begin{align*} \dfrac{2n + 6}{n - 2} \times \dfrac{n - 7}{n - 7} & = \dfrac{(2n + 6)(n - 7)}{(n - 2)(n - 7)} \\ & = \dfrac{2n^2 - 8n - 42}{(n - 2)(n - 7)}\end{align*} $ Now we have: $ = \dfrac{n^2 - 5n + 6}{(n - 7)(n - 2)} - \dfrac{2n^2 - 8n - 42}{(n - 2)(n - 7)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{n^2 - 5n + 6 - (2n^2 - 8n - 42)}{(n - 7)(n - 2)} $ $ = \dfrac{n^2 - 5n + 6 - 2n^2 + 8n + 42}{(n - 7)(n - 2)} $ $ = \dfrac{-n^2 + 3n + 48}{(n - 7)(n - 2)}$ Expand the denominator: $ = \dfrac{-n^2 + 3n + 48}{n^2 - 9n + 14}$